\begin{answer}
    This is trivial from the definition the a covariance matrix for a random variable $X$:
    $$
    Cov(X) = E[(X - E[X])(X - E[X])^T]
    $$

    In this case, $E(X) = 0$ for $X = \nabla_{\theta'}\log p(y;\theta') |_{\theta'=\theta}$. And thus
    $$
    \begin{aligned}
        \mathcal I(\theta) &= Cov_{y\sim p(y;\theta)}[\nabla_{\theta'}\log p(y;\theta')|_{\theta' = \theta}] \\
        &= E_{y\sim p(y;\theta)}[(\nabla_{\theta'}\log p(y;\theta')|_{\theta' = \theta} - 0)(\nabla_{\theta'}\log p(y;\theta')|_{\theta' = \theta} - 0)^T]\\
        &= E_{y\sim p(y;\theta)}[\nabla_{\theta'}\log p(y;\theta')\nabla_{\theta'}\log p(y;\theta')^T|_{\theta' = \theta}]\\
    \end{aligned}
    $$

\end{answer}
